Check sibling questions

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Ex 9.4, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑑𝑦/𝑑π‘₯=(1 βˆ’ cos⁑π‘₯)/(1 + cos⁑π‘₯ ) 𝑑𝑦/𝑑π‘₯ = (1 βˆ’ cos⁑π‘₯)/(1 + cos⁑π‘₯ ) We know that cos 2x = 2cos2 x βˆ’ 1 Putting x = π‘₯/2 cos 2π‘₯/2 = 2 cos2 π‘₯/2 βˆ’ 1 cos x = 2 cos2 π‘₯/2 βˆ’ 1 1 + cos x = 2cos2 π‘₯/2 We know cos 2x = 1 βˆ’ 2sin2 x Putting x = π‘₯/2 cos2 2π‘₯/2 = 1 βˆ’ 2 sin2 π‘₯/2 cos x = 1 βˆ’ 2sin2 π‘₯/2 1 βˆ’ cos x = 2sin2 π‘₯/2 Putting values in equation π’…π’š/𝒅𝒙 = (𝟐 〖〖𝐬𝐒𝐧〗^𝟐 〗⁑〖𝒙/πŸγ€— )/(𝟐 γ€–γ€–πœπ¨π¬γ€—^𝟐 〗⁑〖𝒙/πŸγ€— ) 𝑑𝑦 = (γ€–sin^2 〗⁑〖π‘₯/2γ€— )/cos^(2 )⁑〖π‘₯/2γ€— 𝑑π‘₯ 𝑑𝑦 = tan2 π‘₯/2 𝑑π‘₯ Putting tan2 π‘₯/2 = sec2 π‘₯/2 βˆ’ 1 π’…π’š = ("sec2 " 𝒙/𝟐 " βˆ’ 1" )𝒅𝒙 Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x βˆ’ 1 Putting x = π‘₯/2 tan2 π‘₯/2 = sec2 π‘₯/2 βˆ’ 1 ∫1β–’π’…π’š = ∫1β–’(π’”π’†π’„πŸ 𝒙/πŸβˆ’πŸ) 𝒅𝒙 y = ∫1▒〖𝑠𝑒𝑐2 π‘₯/2 𝑑π‘₯βˆ’ ∫1▒𝑑π‘₯γ€— y = 𝟏/(𝟏/𝟐) tan 𝒙/𝟐 βˆ’ x + C y = 𝟐 tan 𝒙/𝟐 βˆ’ x + C is the general solution (As ∫1▒〖𝑠𝑒𝑐2 (π‘Žπ‘₯) 𝑑π‘₯γ€— = 1/π‘Ž tan ax + C)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.